how to proof ${ NPC \bigcap CO-NPC \ne \varnothing then NP = P ? }$

Question

how proof ${\ \ NPC \ \ \bigcap \ \ CO-NPC \ne \varnothing }$
then ${NP = P ? }$

Answer 1

I don't think you're likely to find any such proof. Given our current level of knowledge, as far as we know it is possible that $\textsf{P} \ne \textsf{NP}$ but $\textsf{NP} = \textsf{co-NP}$ (we cannot prove otherwise). If that were true, then we'd have $\textsf{NPC} = \textsf{co-NPC}$ (and thus $\textsf{NPC} \cap \textsf{co-NPC} \ne \emptyset$) yet $\textsf{P} \ne \textsf{NP}$.

See, e.g., Is the open question NP=co-NP the same as P=NP? and If NP $\neq$ Co-NP then is P $\neq$ NP.